(a) Find the speed at the top of the loop. Write Newton's second law for the...

 

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(a) Find the speed at the top of the loop. Write Newton's second law for the car. At the top of the loop, set n = 0. The force of gravity acts toward the center and provides the centripetal acceleration a = v = v/R. Solve the foregoing equation for 'top' (b) Find the speed at the bottom of the loop Apply conservation of mechanical energy to find the total mechanical energy at the top of the loop. (1) ma = + m m(2 "top top/R) = gR = mg Etop = 1/2mv 2 top + mgh = 1/2mgR+mg(2R) = 2.5mgR Find the total mechanical energy at the bottom of the loop. Energy is conserved, so these two energies may be equated and solved for "bot' Etop = 1/2mv bot 2 1/2mv bot 2.5mgR "bot = 5gR (c) Find the normal force on a passenger at the bottom. (This is the passenger's perceived weight.) Use Equation (1). The net centripetal force is n mg. Solve for n. LEARN MORE m(v2 bot/R) = n - mg n = mg + m(v2 bot/R) =mg+m(5gR/R) = 6mg Remarks The final answer for n shows that the rider experiences a force six times normal weight at the bottom of the loop! Astronauts experience a similar force during space launches. Question Suppose the car goes over a rise at the same speed as part (a), with the same radius of curvature at the peak. What is the normal force as the car passes the top of the rise? 6mg 2mg mg o 0 EXERCISE HINTS: GETTING STARTED | I'M STUCK! A jet traveling at a speed of 1.10 x 102 m/s executes a vertical loop with a radius of 5.25 x 102 m. (See Fig b.) Find the magnitude of the force of the seat on a 70.0 kg pilot at the following positions: (a) the top of the loop 686.7 Draw a free-body diagram of the pilot and note that there are only two forces on him. However, information about the motion of the plane lets us know his acceleration, which then allows us to find the strength of the force of the seat. Note the direction of the acceleration. N (b) the bottom of the loop N (a) Find the speed at the top of the loop. Write Newton's second law for the car. At the top of the loop, set n = 0. The force of gravity acts toward the center and provides the centripetal acceleration a = v = v/R. Solve the foregoing equation for 'top' (b) Find the speed at the bottom of the loop Apply conservation of mechanical energy to find the total mechanical energy at the top of the loop. (1) ma = + m m(2 "top top/R) = gR = mg Etop = 1/2mv 2 top + mgh = 1/2mgR+mg(2R) = 2.5mgR Find the total mechanical energy at the bottom of the loop. Energy is conserved, so these two energies may be equated and solved for "bot' Etop = 1/2mv bot 2 1/2mv bot 2.5mgR "bot = 5gR (c) Find the normal force on a passenger at the bottom. (This is the passenger's perceived weight.) Use Equation (1). The net centripetal force is n mg. Solve for n. LEARN MORE m(v2 bot/R) = n - mg n = mg + m(v2 bot/R) =mg+m(5gR/R) = 6mg Remarks The final answer for n shows that the rider experiences a force six times normal weight at the bottom of the loop! Astronauts experience a similar force during space launches. Question Suppose the car goes over a rise at the same speed as part (a), with the same radius of curvature at the peak. What is the normal force as the car passes the top of the rise? 6mg 2mg mg o 0 EXERCISE HINTS: GETTING STARTED | I'M STUCK! A jet traveling at a speed of 1.10 x 102 m/s executes a vertical loop with a radius of 5.25 x 102 m. (See Fig b.) Find the magnitude of the force of the seat on a 70.0 kg pilot at the following positions: (a) the top of the loop 686.7 Draw a free-body diagram of the pilot and note that there are only two forces on him. However, information about the motion of the plane lets us know his acceleration, which then allows us to find the strength of the force of the seat. Note the direction of the acceleration. N (b) the bottom of the loop N